That's Geophysical Fluid Dynamics.

Adding Rotation

Now that we have the momentum and mass conservation equations, we can modify them to include the effects of rotation (we already have stratification due to the presence of gravity!)

Let us assume that our velocity components \((u,v,w)\) are defined with respect to a fixed reference frame. We would like to calculate \((u_r,v_r,w_r)\), which are the velocities with respect to the rotating reference frame of the spinning Earth.

Assume our reference frame is rotating counter clockwise according to a rotation vector \(\vec{\omega}\). Then our basis vectors \((\hat{i},\hat{j},\hat{k})\) evolve in the following way:

$$
\begin{align*}
\frac{D\hat{i}}{Dt} &= \vec{\omega} \times \hat{i} \\
\frac{D\hat{j}}{Dt} &= \vec{\omega} \times \hat{j} \\
\frac{D\hat{k}}{Dt} &= \vec{\omega} \times \hat{k}
\end{align*}
$$

Then with the help of the product rule, the material derivative of our coordinates \(\vec{x}\) (i.e. the velocity in our inertial frame) is given by

$$\begin{align*}\frac{D\vec{x}}{Dt} &= {\left(\frac{D x}{Dt}\right)}_r\hat{i} + {\left(\frac{D y}{Dt}\right)}_r\hat{j} + {\left(\frac{D z}{Dt}\right)}_r\hat{k} + \vec{\omega} \times \vec{x} \\\\ &= {\vec{v}}_r + \vec{\omega} \times \vec{x}\end{align*}$$

where \(\vec{v}_r\) is the material derivative of \(x\) in the new reference frame.

Lastly, we just need to take one more material derivative to get the accelerations in the
rotating reference frame:

$$\frac{D\vec{v}}{Dt} = {\left(\frac{D\vec{v_r}}{Dt}\right)}_r + 2\vec{\omega} \times \vec{v}_r +\vec{\omega} \times (\vec{\omega} \times \vec{x})$$

Substituting this into the momentum equation (and dropping the subscript so we assume that we
are in the rotating reference frame), we get:

$$\frac{D \vec{v}}{ D t} + 2 \vec{\omega} \times \vec{v} + \vec{\omega} \times (\vec{\omega} \times \vec{x}) = -\frac{\nabla p}{\rho} – g\hat{k}$$

What is \(\vec{\omega}\)?

But what exactly does the rotation vector \(\vec{\omega}\) look like? We know that it should somehow be related to the rotation rate of the Earth. In fact, it should be the projection of the Earth’s rotation vector onto the local basis around whichever point we are interested in.

Say we are in our rotating reference frame, and let \((\hat{i},\hat{j},\hat{k})\) be the local basis vectors in the east, north, and radially outward directions (respectively) around a point on the globe. Then \(\vec{\omega}\) should have no eastward component in this basis.

The rotation rate of the Earth in radians per second is given by:

$$\Omega = \frac{2 \pi}{86400\text{ seconds}} \approx 7.27 \times 10^{-5 }\text{ seconds}^{-1}$$

At any point on the equator, the axis of rotation is parallel to the direction of north, so
\(\vec{\omega} = (0,\Omega,0)\). At higher latitudes, the rotational component
gets stronger in the \(\hat{k}\) direction and weaker in the \(\hat{j}\) direction. In fact, letting \(\varphi\) represent latitude,

$$\vec{\omega}(\varphi) = (0,\Omega\cos(\varphi),\Omega\sin(\varphi))$$

So

$$\vec{\omega} \times \vec{v} = \Omega (w \cos(\varphi) – v\sin(\varphi), u\sin(\varphi), -u\cos(\varphi))$$

There are a couple of more things we must do to arrive at our final equations. These are all
approximations, and while they are good approximations, they will lead to an expression
of momentum conservation that is no longer exact.

• Because \(\Omega\) is small in magnitude, \(|\Omega^2|\) is really small. Thus, we typically neglect
  the term \(\vec{\omega} \times (\vec{\omega} \times \vec{x})\) in the momentum equation because its magnitude is insignificant compared with the other terms.
• We will use the fact that gravity is much stronger than the vertical component of rotation to get rid of its contributions to the evolution of vertical velocities. Thus, we will only consider the \(\hat{i}\) and \(\hat{j}\) components of \(\vec{\omega} \times \vec{v}\).
• Because vertical velocities are much smaller than horizontal velocities on Earth, we can discard the term \(w \cos(\varphi)\) in the evolution equation for \(u\), as its magnitude is much smaller than \(|v\sin(\varphi)|\)

Lastly, let \(f = 2 \Omega \sin(\varphi)\). This is called the Coriolis parameter and it will come up time and time again. Then our momentum equations are:

$$\frac{D \vec{v}}{ D t} + f\hat{k} \times \vec{u} = -\frac{\nabla p}{\rho} – g\hat{k}$$

where \(\vec{u} = (u,v,0)\) is the vector of horizontal velocities. Stated in terms of individual velocities,

$$
\begin{align*}
\frac{D u}{ D t} – fv &= -\frac{1}{\rho}\frac{\partial p}{\partial x} \\\\
\frac{D v}{ D t} + fu &= -\frac{1}{\rho}\frac{\partial p}{\partial y} \\\\
\frac{D w}{ D t} &= -\frac{1}{\rho}\frac{\partial p}{\partial z} – g
\end{align*}
$$

Hydrostatic Balance

One last thing: Notice that we have one mass conservation equation and three momentum equations, giving us four equations total. But we have five variables: \(u\), \(v\), \(w\), \(p\), and \(\rho\). What do we do? One thing we have not included is an equation of state, which describes a governing relationship between our variables. The choice of equation depends on the nature of the fluid we’re looking at.

But in this case, we won’t need one! One important property of geophysical flows is that horizontal velocity scales are often much larger than vertical ones. If we assume that vertical motion is small, then we may eliminate \(w\) and its material derivative from the equations. Then we are left with only four variables! Our vertical momentum equation becomes:

$$\frac{\partial p}{\partial z} = -\rho g$$

This is known as hydrostatic balance. The density controls the vertical structure of pressure, but there is no motion between the different z-levels. This inhibition of vertical motion is unrealistic at smaller scales, but often works nicely for large scale geophysical flows.

Summary

After all that work, we have arrived at the mass conservation and momentum equations for a geophysical system in hydrostatic balance! These equations combine both rotation and stratification – two key components of GFD!

\begin{align*}
\frac{D \rho}{D t} + \rho \nabla \cdot \vec{u} &= 0 \\\\
\frac{D u}{ D t} – fv &= -\frac{1}{\rho}\frac{\partial p}{\partial x} \\\\
\frac{D v}{ D t} + fu &= -\frac{1}{\rho}\frac{\partial p}{\partial y} \\\\
\frac{\partial p}{\partial z} &= -\rho g
\end{align*}